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CHEM 3P40 (2)
Lecture 2

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Department
Chemistry
Course
CHEM 3P40
Professor
Costa Metallinos
Semester
Fall

Description
Lecture 2: Determination od the Molecular Formula of an Unknown Friday, September 06, 2013  Elemental analysis: complete combustion type C H O →xO y z O 2 2  Use the number of CO and2H O to 2et the molecular formula Example: 100mg of pure compound yields 275mg of CO and 225mg o2 H O. What is the 2olecular formula of the compound? Take the masses of each CO and2H ) and2divide them by their respective molecular weights: For CO : 275mg/(44mg.mmol )= 6.25mmol  2 -1  For H 2: 225mg/(18mg/mmol )= 12.5mmol Calculate the number of moles of both C and H atoms, and divide by the smallest number of moles  In CO 2here is only 1 carbon atom, therefore: 6.25*1 = 6.25  In H2O there are 2 atoms of H, therefore: 12.5*2 = 25  6.25mmol < 25mmol  For C: 6.25/6.25 = 1  For H: 25/6.25 = 4  Therefore we get the following empirical formula: CH 4 Is this the empirical formula or the molecular formula?  Can you double CH ? (4H )*2 →4 H NOT P2SS8BLE!  In this case, the empirical formula is also the molecular formula! What about the number of O?  Calculate the mass of C and H, then see if it accounts for the original mass.  For C: 6.25mmol * (12.011mg*mmol ) = 75mg1 -1  For H: 25mmol * (1.0079mg*mmol ) = 25mg  Total mass: 75mg+25mg = 100mg  There are no O in the compound  The empirical formula is indeed CH 4 Example: In a given sample, there is 52.2% C and 13.0% H (<100%). What is the molecular formula of the compound? Assume the remaining percentage is O  100-(52.2+13) = 34.8 Calculate the number of moles of C, H and O in 100mg of compound by dividing the mass by the atomic weight -1  For C: 52.2g/(12.0g*mol ) = 4.35mol For H: 13g/(1g*mol ) - 13mol  -1  For O: 34.8g/(16g*mol ) = 2.175mol Normalize by dividing by the smallest integer: 2.175<4.35<13  For C: 4.35/2.175 = 2 For H: 13/2.175 = 5.977 (round up to 6)   For O: 2.175/2.175 = 1  Therefore we have the following empirical formula: C H O 2 6 In order to get the molecular formula, we need the molecular weight of the compound (using MS).  Rule of 13 If only the molecular weight is known for a given compound, a possible formula can be derived by dividing the molecular weight into CH units (mass of 13g). The remainder of the molecular weight is attributed to H. Example: a compound has a molecular weight of 66g 66 = (5*13)+1 Therefore, we have 5 CH units and 1 H The formula for the compound is: C H 5 6 Another possible formula can be found by adding heteroatoms and subtracting CH units n Example: Mass = 16g O or CH 4 Mass = 14g N or CH 2 Alternative molecular formulas:  C H + O - CH = C H O 5 6 4 4 2  C 5 +6N - CH = C2H N 4 4  Unsaturation Number (UN) The unsaturation number of an organic compound gives the number of double bonds and/or rings. General Organic Compound:
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